Canonical group structure on hom-sets in an additive category I am trying to understand why a given category (without any extra structure) is either abelian or not abelian i.e. why the definition can be expressed in such a way that there is no need to add any further structure for 'abelianness' to make sense. One or two difficulties have arisen in making sense of the canonical group structure on the hom-sets.
A category is said to be semiadditive if it has a zero object $0$, and where each pair of objects $A$ and $B$ have a biproduct $A\oplus B$. We can then construct an addition of morphisms $f,g:A\to B$ in such a way that $\text{Hom}(A,B)$ is a commutative monoid and composition is biadditive, although it would seem that there are multiple choices for this addition.
A semiadditive category is additive if each $f:A\to B$ has an inverse $-f:A\to B$, so that $\text{Hom}(A,B)$ is an abelian group. According to Wikipedia, the 'two additions' on the hom-sets in an additive category must agree. What explicitly are these two different additions? My best guess would be the following:

To elaborate: Combining two copies of the identity morphism $1_A:A\to A$ gives a diagonal $\Delta:A\to A\oplus A$. Similarly, we get a codiagonal $\nabla:B\oplus B\to B$. For $f\oplus g$, we can either first find a morphism $A\oplus A\to B$ and then get $A\oplus A\to B\oplus B$, or  alternatively we can first do $A\to B\oplus B$, and then $A\oplus A\to B\oplus B$. There are other construction with matrices of morphisms, but I'm not sure if they are relevant here.
In short, I need to understand precisely which morphisms are being referred to and where the uniqueness comes from.
 A: Uniqueness holds with weaker hypotheses than semi-additivivty. It suffices to have a category with zero morphisms for which square projections are jointly epimorphic or cosquare injections are jointly monomorphic. Replacing squares and cosquares with products or corpoduct yields the hypotheses needed to have a matrix calculus on the category, so the notion of matrices of morphisms is not unrelated.
Let me explain the former case of jointly epimorphic kernels of projections.
Fix an object $C$ that has a square with projections $\pi_i\colon C^2\to C$ for $i=1,2$, so that any pair of morphisms $f_1,f_2\colon B\to C$ correspond to a unique morphism $(f_1,f_2)\colon B\to C^2$ such that $\pi_i(f_1,f_2)=f_i$. Recall that such an association is natural: for any morphism $h\colon A\to B$, we have $\pi_i((f_1,f_2)h)=f_ih$, whence $(f_1,f_2)h=(f_1h,f_2h)$.
Consequently, any morphism $*\colon C^2\to C$ determines a binary operation on morphisms $\mathrm{Hom}(B,C)$ given by $f_1*f_2=*(f_1,f_2)$ that is natural in the sense that for any $h\colon A\to B$, $(f_1*f_2)h=*(f_1,f_2)h=*(f_1h,f_2h)=(f_1h)*(f_2h)$. Moreover, this association is a bijection since $\pi_1*\pi_2=*(\pi_1,\pi_2)=*\mathrm{id}_{C^2}=*$ and $f_1*f_2=\pi_1*\pi_2(f_1,f_2)$.
Thus, uniqueness of natural binary operations on $C$ is the same as uniqueness of morphisms $C^2\to C$.
We can do even better for natural unital binary operations in a category with zero morphisms. These are natural binary operations on $\mathrm{Hom}(B,C)$ with a chosen unit $0\colon B\to C$ such that for any $h\colon A\to B$, we have $0h=0$. In particular, we must have an endomorphism $0\colon C\to C$ such that $0e=0$ for any other endomorphism $e\colon C\to C$. In particular, if the category has zero morphisms, i.e. a choice of $0\colon A\to B$ for each pair of objects $A$ and $B$ such that $f0=0$ and $0g=0$, then the unit of any natural binary operation has to be the zero morphism.
We now have morphisms $i_1=(\mathrm{id}_C,0),i_2=(0,\mathrm{id}_C)\colon C\to C^2$, that are monomorphisms because $\pi_ji_j=\mathrm{id}_C$, and, in the case where the category has zero morphisms, $i_j$ are also the kernels of $\pi_{3-j}$ for $j=1,2$, i.e. $\pi_{3-j}h=0$ means $h=(h_1,h_2)$ satisfies $h_{3-j}=0$, and hence factors (uniquely) as $i_jh_j$.
These morphisms are significant because $0\colon C\to C$ is a unit for a natural binary operation corresponding to $*\colon C^2\to C$ if and only if $*i_j=\mathrm{id}_C$ for $j=1,2$. In particular, for a category with zero morphisms this condition is implied by the kernels of $\pi_j$ being jointly epimorphic, i.e. such that $fi_j=gi_j$ for $j=1,2$ implies $f=g$.
The case of $C$ having a bipower in a category with zero morphisms is by definition the one in which $i_j\colon C\to C^2$ are a coproduct cocone, hence in particular jointly epimorphic, in which case the folding map $\nabla\colon C^2\to C$ is by definition one giving rise to the (necessarily unique) natural unital binary operation on $\mathrm{Hom}(A,C)$.

Going beyond the original question, if the category has a matrix calculus (e.g. zero morphisms and products with jointly epimorphic projections), then the natural unital binary operation (if it exists) is in fact a natural commutative monoid. It is then a natural abelian group if and only if the cone obtained from $*\colon C^2\to C$ combined with one of the projection morphisms $C^2\to C$ is also a product cone.
